Banking Exams Preparation, Bank PO/Clerical Entrance India, Banking Career, Ask Banking Experts

ashwani kumar

A ball falls on an inclined plane of inclination 45 degree. from a height 5m above the point of impact and makes an inelastic collision. Where will it hit the plane.
Where coeffient of restitution e=3/4.

Akshay Bhimrajka

velocity accquired just before impact is 10m/s
since it is not elastic,
using conservation of momentum and newton's law of collision,
we get velocity in direction of down the plane = 15/(2 * root 2)
and velocity perpendicular to the plane = 5 * root 2
and the angle wrt the surface of the plane is 37
using this we can find range which turns out to be 18.5675 m...

ashwani kumar

Answer provided by akshax is correct but i am still confused.
Not able to fint velocity componenets ...
Plzz anybody explain it....

ashwani kumar

Anybody plz solve this prob its urgent......

Priyesh

see the figure attached(click to enlarge)

just before striking velocity of ball velocity of ball = root(2g5) = 10m/sec

let the ball rebound at a speed v making an angle

with the plane

now line of impact is the dotted one as shown

so velocity of seperation alng line of impact = 3/4(velocity of approach along line of impact)

=>v sin

= 3/4 * 10cos45 (u = 10) --------------------1

now momentum can be conserved perpendicular to the line of impact

so musin45 = mvcos

(u =10)

=> vcos

= 10cos45 (as sin45 = cos45) ---------------------------------------2

diving 1 by 2

tan

= 3/4

=>

= 37degress ( value of

not needed but just to explain wat akshax did)

now g along incline is - g/root(2)

perpendicular to incline is g/root(2)

so for vertical(perpendicular to incline)

0 = vsin

- gsin

t

0 = 15/2root(2) - g/root(2)t => t = 3/4 so total time in again reaching ground is 2t = 3/2 = 1.5 sec

so

now for horizontal(along the plane)

dist. travelled in 1.5 sec = range = 10/

2 (1.5) + 1/2(g/

2)(1.5)2 = 18.564 meters

joy francis

The velocity of the ball when it hits the inclined plane = root (2*g*5)=10m/s in vertically downward direction.

.: It's velocity about the common normal direction = 10 (cos45) = 10/

2 m/s

e = vsep / vapp in CN direction

e = vsep / (10/

2)

since e = 3/4 ..

v = 15/2

2 m/s along the CN direction.....i.e the direction perpendicular to the plane

and since no force is applied in the direction of incline the vel of separation along the incline would not change. So it will remain 10/

2.

Now just use properties of a projectile on inclined plane to get the answer....

g along incline = g/

2

g perpendicular to incline = g/

2

Y = u_yt+1/2(a_yt²)

0 = 15/2

2 - gt/2

2

=> t = 1.5 sec.

X = u_xt+1/2(a_xt²)

= 10/

2 (1.5) + 1/2(g/

2)(1.5)2

= 18.564 meters.......(ans)

ashwani kumar

Thanks Priyesh and joyfrancis

Ask & Discuss Questions with Community & Experts